Question: Compute $\cos \left( \arcsin \frac{5}{13} \right).$
Solution: Consider a right triangle where the opposite side is 5 and the hypotenuse is 13.

[asy]
unitsize (0.3 cm);

draw((0,0)--(12,0)--(12,5)--cycle);

label("$12$", (6,0), S);
label("$13$", (6,5/2), NW);
label("$5$", (12,5/2), E);
label("$\theta$", (5,1));
[/asy]

Then $\sin \theta = \frac{5}{13},$ so $\theta = \arcsin \frac{5}{13}.$  By Pythagoras, the adjacent side is 12, so $\cos \theta = \boxed{\frac{12}{13}}.$